How can school children intuit why over 100, D is larger? But under 100, D% is larger?
I can prove the Rule of 100 algebraically, below. But my school kids are hankering after intuition, and a plainer explanation.
Follow the Rule of 100
Should discounts be percentages or absolutes?
Consider a \$150 blender. Should you offer 20% off? Or an equivalent \$30 off?
Answer:
- **Over \$100?** Give absolutes (e.g., \$30)
- Under \$100? Give percents (e.g., 20%)
In both cases, you show the higher numeral. For a \$50 blender, 20% off is the same as \\$10 off — yet 20% is more persuasive because it’s a higher numeral. For a \$150 blender, the absolute discount (\\$30 off) is a higher numeral (González, Esteva, Roggeveen, & Grewal, 2016).
References
González, E. M., Esteva, E., Roggeveen, A. L., & Grewal, D. (2016). Amount off versus percentage off—when does it matter?. Journal of Business Research, 69(3), 1022-1027.
Let d = discount, p = price. Then $d, p > 0$ because there is no free lunch and the quotation is expatiating on discounts. Then
\$ off vs. % off $\iff p - d \quad \text{ vs } \quad p - (d/100)p \iff -d \quad \text{ vs } \quad -dp/100 \iff 1 \quad \text{ vs } \quad p/100 \iff p \quad \text{ vs } \quad 100. $
2 answers
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If cost is $c$, sold for $s$, discount, $d = c - s$
Discount percentage = $\frac{c - s}{c} \times 100 = (c - s)\times \frac{100}{c}$
Absolute discount = $c - s$
As you can see ...
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When $c < 100$, $\frac{100}{c} >1$, and so, $(c - s)\times \frac{100}{c} > (c - s)$. The percentage discount > The absolute discount.
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When $c > 100$, $\frac{100}{c} < 1$, and so $(c - s) \times \frac{100}{c} < (c -s)$. The absolute discount > The percentage discount.
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When $c = 100$, The absolute discount = The percentage discount. $\frac{100}{c} = \frac{100}{100} = 1$
EDIT 1 START
Intuitively, we begin where it starts (tautologies are always so out there), which is Discount (as a) percentage (of cost price) = $\frac{c - s}{c} \times 100$, where $c$ = cost price and $s$ is the selling price.
Notice that $\frac{c - s}{c} \times 100 = (c - s) \times \frac{100}{c}$. What we've achieved by doing this is that we can compare how the discount percentage varies with how $c$ relates to $100$.
When $c = 100$, we get $c - s$ which is the absolute discount amount, which is also equal to the discount percentage.
When $c < 100$,$c - s$ gets multiplied by an improper fraction $\frac{100}{c} > 1$, which means the absolute discount amount $c - s$ will be less than the percentage discount $\frac{c - s}{c} \times 100$ which we saw is the same as $(c - s) \times \frac{100}{c}$.
When $c > 100$, $c - s$ gets multiplied by an proper fraction $\frac{100}{c} < 1$, which means the absolute discount amount $c - s$ will be greater than the percentage discount $\frac{c - s}{s} \times 100$, which we say is the same as $(c - s) \times \frac{100}{c}$.
EDIT 1 END
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(I’m going to do this post in pounds so that I don’t have to escape dollar signs everywhere. But the currency doesn’t matter.)
“Per cent” is, literally, “for each hundred”. Imagine making a literal pile of money from the cost of the item, and splitting it up into stacks of £100. So if the item costs £500, you have five stacks.
Now, a discount that’s “ten pounds” means just that: take £10 out of the pile (any stack, it doesn’t matter) and put it back in your wallet.
But “ten per cent” means £10 out of every single one of those stacks. For five stacks, that’s £50 I get to keep! I’d rather have that discount.
What if the price is only £100, though? In that case, there’s only one stack, and “£10 from one stack” and “£10 from every stack” mean the same thing.
Below £100, the intuition is a little bit trickier, because you have to imagine the 10% case as taking “part of” £10 from a stack that’s “part of” £100. But it’s not an insurmountable hurdle to envision that, and grasp that it won’t be as good as taking out a full £10.
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